ln(x^2+1) / (x^2+1) の積分

Problem / 問題

Show that: / 以下を示せ：
$$\int_0^\infty \frac{\ln (x^2+1)}{x^2+1} dx = \pi \ln 2$$

Solution / 解答

Let $x = \tan \theta$, $dx = \sec^2 \theta d\theta$.

$$\int_0^\infty \frac{\ln (x^2+1)}{x^2+1} dx = \int_0^{\pi/2} \frac{\ln (\sec^2\theta)}{\sec^2 \theta} \sec^2 \theta d\theta = -2\int_0^{\pi/2} \ln (\cos \theta) d\theta $$

Let $I = \int_0^{\pi/2} \ln (\cos \theta) d\theta$. Then $\theta \to \frac{\pi}{2}-\theta$ gives $I = \int_0^{\pi/2} \ln (\sin \theta) d\theta$.

Hence $$2I + \frac{\pi}{2} \ln 2 = \int_0^{\pi/2} \left(\ln (\cos \theta) +\ln (\sin \theta) + \ln 2\right) d\theta = \int_0^{\pi/2} \ln (\sin 2\theta) d\theta$$

By symmetry, $\int_0^{\pi/2} \ln (\sin 2\theta) d\theta = 2 \int_0^{\pi/4} \ln (\sin 2\theta) d\theta = \int_0^{\pi/2} \ln(\sin\phi)d\phi = I$, where $\phi = 2\theta$. Hence $2I+\frac{\pi}{2}\ln 2 = I$; $I = -\frac{\pi}{2}\ln 2$; $\int_0^\infty \frac{\ln (x^2+1)}{x^2+1} dx = \pi \ln 2$.